For non-zero finite output, it must be that
\[p z_š = w = z_š½\]
and for zero output of š, it must be that
\[w \geq p z_š\]
In either case, profits will be zero.

Solution to consumer's problem:

In an interior solution where \(š>0\),
\[\frac{(š+1)}{\gamma š½} = p\]
\[ pš + š½ = w \]
\[š½ = \frac{w+p}{\left(1+\gamma\right)}\]
\[š = \frac{w\gamma-p}{p\left(1+\gamma\right)}\]
And in a solution where \(š=0\),
\[š½ = w\]
Note that interior solution requires that \(p < w\gamma\)

Click to expand work for consumer problem
Assume profits are zero. Will be true in equilibrium, so no big deal. Then
Assuming interior solution where \(š½,š>0\), Lagrangian is:
\[\mathcal{L} = \ln (š½) + \gamma \ln (š+1) - \lambda [ pš + š½ - w ] \]
FOCs:
\[{1 \over š½} = \lambda\]
\[{\gamma \over (š+1)} = \lambda p\]
\[pš + š½ = w \]
Implies
\[{\gamma \over (š+1)} = {1 \over š½} p\]
Solve this to express each variable in terms of the other:
\[š½=š\frac{p}{\gamma}+\frac{p}{\gamma}\]
\[š=š½\frac{\gamma}{p}-1\]
Combine these with budget constraint to solve explicitly for allocation variables:
\[š½ = \frac{w+p}{\left(1+\gamma\right)}\]
\[š = \frac{w\gamma-p}{p\left(1+\gamma\right)}\]

Combine problems to get equilibrium

If \(\gamma z_š > 1\), then we have the interior solution:
\[š½ = z_š½ \cdot \left( \frac{1+z_š}{z_š\left(1+\gamma\right)} \right) \]
\[š = \frac{\gamma z_š - 1}{\left(1+\gamma\right)}\]
\[N_š½ = \frac{š½}{z_{š½}} = \frac{\left(z_{š}+1\right)}{z_{š}\left(1+\gamma\right)}\]
\[N_{š} = \frac{š}{z_{š}}=\frac{\gamma-\frac{1}{z_{š}}}{1+\gamma}\]
Otherwise, if \(\gamma z_š \leq 1\),
\[N_š½ = 1\]
\[N_š = 0\]
\[š½ = w = z_š½\]

Click to expand work for solving for the equilibrium allocations
For an interior solution, we have the following characterizing equations:
From consumer:
\[\frac{(š+1)}{\gamma š½} = p\]
\[ pš + š½ = w \]
\[š½ = \frac{w+p}{\left(1+\gamma\right)}\]
\[š = \frac{w\gamma-p}{p\left(1+\gamma\right)}\]
\[š½,š>0\]
From firms:
\[p z_š = w = z_š½\]
From Market Clearing
\[N_š + N_š½ = 1\]
\[ Y_š = z_š N_š \]
\[ Y_š½ = z_š½ N_š½ \]
Note how this implies that
\[p=\frac{ z_š½ }{ z_š }\]
And substitute out the prices to get allocations:
\[š½ = \frac{ z_š½ + {z_š½ \over z_š} }{\left(1+\gamma\right)}
= z_š½ \frac{1+z_š}{z_š\left(1+\gamma\right)} \]
\[š = \frac{z_š½ \gamma-{z_š½ \over z_š}}{{z_š½ \over z_š}\left(1+\gamma\right)}
= \frac{\gamma z_š - 1}{\left(1+\gamma\right)}\]
\[N_š½ = \frac{š½}{z_{š½}} = \frac{\left(z_{š}+1\right)}{z_{š}\left(1+\gamma\right)}\]
\[N_{š} = \frac{š}{z_{š}}=\frac{\gamma-\frac{1}{z_{š}}}{1+\gamma}\]
Note that the above equations are consistent with \(š>0\) iff \(\gamma z_š > 1\). If this isn't the case, then it must instead be that \(š=0\) and
\[N_š½ = 1\]
\[N_š = 0\]
\[š½ = w = z_š½\]